[Leetcode 1025] Divisor Game

原题说明

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there is a number N on the chalkboard. On each player’s turn, that player makes a move consisting of:

Choosing any x with 0 < x < N and N % x == 0.
Replacing the number N on the chalkboard with N - x.
Also, if a player cannot make a move, they lose the game.

Return True if and only if Alice wins the game, assuming both players play optimally.

Example 1:
Input: 2
Output: true
Explanation: Alice chooses 1, and Bob has no more moves.

Example 2:
Input: 3
Output: false
Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves.

Note:

1 <= N <= 1000

解题思路

如果N为偶数，则坚持选择x = 1，对方得到N - 1为大于0的奇数
如果N为奇数：

如果N > 1, 其因子必然也是奇数，所以无论选择任何x，N - x都将变为偶数
由于x < N，给予对方的N - x必然是 > 0的偶数
如果N = 1，直接失败。
综上，N为偶数时必然大于0，只需要坚持选择x = 1，最终对方会得到N = 1而失败。
所以N为偶数时返回true。

用Flip Game II的方法也可以。

示例代码 (cpp)

class Solution {
public:
    bool divisorGame(int N) {
        return N % 2 == 0;
    }
};

示例代码 (java)

class Solution {
    public boolean divisorGame(int N) {
        return N % 2 == 0;
    }
}

示例代码 (python)

class Solution(object):
    def divisorGame(self, N):
        """
        :type N: int
        :rtype: bool
        """
        return N % 2 == 0

复杂度分析

时间复杂度: O(1)
空间复杂度: O(1)

归纳总结

我们在Youtube上更新了视频讲解，欢迎关注！